Question: $\dfrac{ 8x + 8y }{ -5 } = \dfrac{ 10x + 5z }{ 6 }$ Solve for $x$.
Multiply both sides by the left denominator. $\dfrac{ 8x + 8y }{ -{5} } = \dfrac{ 10x + 5z }{ 6 }$ $-{5} \cdot \dfrac{ 8x + 8y }{ -{5} } = -{5} \cdot \dfrac{ 10x + 5z }{ 6 }$ $8x + 8y = -{5} \cdot \dfrac { 10x + 5z }{ 6 }$ Multiply both sides by the right denominator. $8x + 8y = -5 \cdot \dfrac{ 10x + 5z }{ {6} }$ ${6} \cdot \left( 8x + 8y \right) = {6} \cdot -5 \cdot \dfrac{ 10x + 5z }{ {6} }$ ${6} \cdot \left( 8x + 8y \right) = -5 \cdot \left( 10x + 5z \right)$ Distribute both sides ${6} \cdot \left( 8x + 8y \right) = -{5} \cdot \left( 10x + 5z \right)$ ${48}x + {48}y = -{50}x - {25}z$ Combine $x$ terms on the left. ${48x} + 48y = -{50x} - 25z$ ${98x} + 48y = -25z$ Move the $y$ term to the right. $98x + {48y} = -25z$ $98x = -25z - {48y}$ Isolate $x$ by dividing both sides by its coefficient. ${98}x = -25z - 48y$ $x = \dfrac{ -25z - 48y }{ {98} }$